6 Answers
6

$\displaystyle \int_a^b f(x) dx$ is the area under the curve $y=f(x)$ in the interval $(a,b)$ when you integrate from left to right.
$\displaystyle \int_a^b f(a+b-x) dx$ is the area under the curve $y=f(x)$ in the interval $(a,b)$ when you integrate from right to left.
Hence, both are equal.

For any point $x$ in the range $x=a$ to $x=b$
we can define a variable scalar $P$ such that $x$ divides the interval $a...b$ in the ratio $(P)$:$(1-P)$ where $0 <= P <=1$.

Now we can define any point $x$ in two ways:
$x1 = a + P(b-a)$ and
$x2 = b - (1-P)(b-a)$

Now let us insert $x2$, $x1$ into the two different functions $G1$,$G2$

$G1(x) = F([x2]) = F([b - (1-P)(b-a)]) = F(a + P(b-a))$

$G2(x) = F(a+b-[x1]) = F(a + b - [a + P(b-a)]) = F(b - P(b-a))$

therefore
$\int_0^1G1(x)\,\mathrm{d}P =\int_0^1G2(x)\,\mathrm{d}P$
because in the former integration we move across the interval from $x=a$ to $x=b$ and in the latter integration we move across the same interval from $x=b$ to $x=a$.